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The language of motion: position, displacement, distance, velocity, speed and acceleration, and which of them are vectors. It comes up in short questions on average speed and average velocity, and in problems where position and velocity are given as \(\mathbf{i}\), \(\mathbf{j}\) vectors.
Key facts
Vectors: displacement, velocity, acceleration. Scalars: distance, speed, time
Average speed = total distance ÷ total time
Average velocity = total displacement ÷ total time
A vector has size and direction; a scalar has size only.
Position: where an object is, relative to a fixed origin \(O\).Displacement: the change in position (a vector).
Distance travelled: the total length of the path (a scalar). It can never be negative.
Velocity: rate of change of displacement (a vector). Speed: the magnitude of the velocity (a scalar, never negative).
Acceleration: rate of change of velocity (a vector). An acceleration opposite to the direction of motion slows the object down: a deceleration or retardation.
In a straight line, a sign shows the direction. Choose a positive direction and stick to it: a velocity of −3 m s−1 means a speed of 3 m s−1 in the negative direction.
Averages
diagram
Average speed = total distance travelled ÷ total time taken.
Average velocity = total displacement ÷ total time taken. It is a vector, so give its direction.
Distance counts every metre; displacement is just start to finish
After a round trip the displacement is zero, so the average velocity is zero, but the average speed is not.
An object moving at constant speed on a curved path has a changing velocity, because its direction changes.
Vectors in two dimensions
diagram
\(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors (often east and north). A position vector is \(\mathbf{r} = x\mathbf{i} + y\mathbf{j}\)
For a velocity \(\mathbf{v} = a\mathbf{i} + b\mathbf{j}\), the speed is \(|\mathbf{v}| = \sqrt{a^2 + b^2}\), and the angle with \(\mathbf{i}\) comes from \(\tan\theta = \dfrac{b}{a}\). Sketch the vector to get the right quadrant.
v = 3i − 4j: speed √(3² + 4²) = 5, at tan−1(4/3) = 53.1° below i
Constant velocity: \(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\), where \(\mathbf{r}_0\) is the position when \(t = 0\)
Displacement from \(A\) to \(B\):\(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\). The distance \(AB\) is its magnitude.
With \(\mathbf{i}\) east and \(\mathbf{j}\) north, 'due north of \(O\)' means the \(\mathbf{i}\) component is zero (and the \(\mathbf{j}\) component positive); 'due east of \(O\)' means the \(\mathbf{j}\) component is zero (and the \(\mathbf{i}\) component positive).
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Distance, displacement and averages
3 to 5 marksGrade E
Draw a quick line diagram and mark a positive direction.
Distance: add the lengths of all the parts. Displacement: the final position relative to the start, with a direction.
Divide by the total time, including any time spent at rest.
Example. A cyclist rides 600 m due east in 100 s and then rides 200 m due west in 60 s. (a) Find the total distance travelled. (b) Find the average speed of the cyclist. (c) Find the average velocity of the cyclist.
Show the model answerHide the model answer
(a) 600 + 200 = 800 m
(B1)
(b) \(\dfrac{800}{160}\) (M1) = 5 m s−1
(A1)
(c) Displacement = 600 − 200 = 400 m due east (M1); average velocity = \(\dfrac{400}{160}\) = 2.5 m s−1 due east
(A1)
Speed and position with a constant velocity vector
4 to 6 marksGrade C
Speed is the magnitude of the velocity vector.
Write the position at time \(t\) as \(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\), collecting the \(\mathbf{i}\) and \(\mathbf{j}\) terms.
Turn the condition into an equation, e.g. 'due east of \(O\)' means the \(\mathbf{j}\) component is zero.
Solve for \(t\) and check that it makes sense.
Example. [\(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors due east and due north.] A boat moves with constant velocity \((3\mathbf{i} - 4\mathbf{j})\) m s−1. At time \(t = 0\) it is at the point with position vector \((2\mathbf{i} + 10\mathbf{j})\) m relative to a fixed origin \(O\). (a) Find the speed of the boat. (b) Find the position vector of the boat at time \(t\) seconds. (c) Find the value of \(t\) when the boat is due east of \(O\)
(c) \(10 - 4t = 0\) (M1), so \(t = 2.5\) (A1). The \(\mathbf{i}\) component is then 9.5, which is positive, so the boat is east of \(O\).
Two objects with constant velocities: do they meet?
5 to 7 marksGrade A
Write the position vector of each object at time \(t\), using the same origin and the same \(t\)
They meet only if both the \(\mathbf{i}\) and the \(\mathbf{j}\) components are equal at the same time.
Solve using one component, then check that value of \(t\) in the other.
Finish with a conclusion giving the time and the position.
Example. [\(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors due east and due north.] At noon, ship \(P\) is at the point with position vector \((4\mathbf{i} + 2\mathbf{j})\) km and ship \(Q\) is at the point with position vector \((10\mathbf{i} - 4\mathbf{j})\) km. \(P\) moves with constant velocity \((2\mathbf{i} + 3\mathbf{j})\) km h−1 and \(Q\) moves with constant velocity \((-\mathbf{i} + 6\mathbf{j})\) km h−1. Show that, if neither ship changes its velocity, the ships will collide, and find the time and the position vector of the point where this happens.
Sketch a vector before finding an angle, so you know which quadrant it is in.
Sense check: the average speed can never be less than the magnitude of the average velocity.
Exam technique
Give a direction with every velocity or displacement, e.g. '2.5 m s−1 due east'.
Keep vector answers in \(\mathbf{i}\), \(\mathbf{j}\) form unless a magnitude is asked for, and underline vectors in handwriting.
In 'show that' questions write every substitution and finish with a clear statement.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Distinguish distance from displacementDisplacement is the vector from the start to the finish; distance is the total length of the path.
Grade E
Find average speed and average velocityDivide the total distance (for speed) or the total displacement (for velocity) by the total time.
Grade D
Use signs to show direction in one dimensionChoose a positive direction; a negative velocity means moving the opposite way at that speed.
Grade C
Find speed and direction from a velocity vectorSpeed is \(\sqrt{a^2 + b^2}\) for \(a\mathbf{i} + b\mathbf{j}\); find the direction using \(\tan\theta\) and a sketch.
Grade B
Use \(\mathbf{r} = \mathbf{r}_0 + \mathbf{v}t\) for constant velocityWrite the position vector at time \(t\) and use it to answer questions about where the object is.
Grade A
Solve problems with two moving objectsWrite both position vectors in terms of \(t\), then equate components or use the vector between them.
Quick recall
Cover the answers and test yourself. The app has these as flashcards that come back just before you'd forget them.
A car is driven once around a closed circuit of length 1.2 km and returns to its starting point. State the displacement of the car from its starting point at the end of the lap.
Zero (0 m) m
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy6 marks
(a) Which two of the following quantities are vectors? Tick (✓) two boxes.[2]
speed
displacement
distance
velocity
time
(b) A car is driven once around a closed circuit of length 1.2 km and returns to its starting point. State the displacement of the car from its starting point at the end of the lap.[1]
(c) The lap takes 80 seconds. Find the average speed of the car, giving your answer in m s−1.[2]
(d) Explain why the average velocity of the car for the lap is zero, even though its average speed is not.[1]
Show the answer and mark scheme
(a)Answer: displacement; velocity
(b)Answer: Zero (0 m) m
B1 for zero
Worked solution: The car finishes where it started, so its displacement from the starting point is zero, even though it has travelled 1.2 km.
(c)Answer: \(15\) m s−1
M1 for \(\frac{1200}{80}\)
A1 for 15 (m s−1)
Worked solution: \(\text{average speed} = \frac{\text{distance}}{\text{time}} = \frac{1200}{80} = 15\text{ m s}^{-1}\)
(d)Answer: Average velocity is displacement ÷ time, and the displacement for the lap is zero; average speed uses the distance travelled, 1200 m.
B1 for average velocity uses displacement (zero), whereas average speed uses distance travelled
Worked solution: Average velocity = displacement ÷ time = 0 ÷ 80 = 0. Average speed = distance ÷ time uses the 1200 m actually travelled.
Question 2Medium6 marks
A cyclist moves along a straight path which lies in the east–west direction. Starting from a point \(A\), the cyclist rides 1440 m due east in 4 minutes, and then rides 900 m due west in 2 minutes 30 seconds, finishing at the point \(B\).
(a) Find the average speed of the cyclist for the whole journey.[2]
(b) Find the magnitude and direction of the average velocity of the cyclist for the whole journey, giving the magnitude to 3 significant figures.[3]
(c) Explain why the average speed is greater than the magnitude of the average velocity.[1]
Show the answer and mark scheme
(a)Answer: \(6\) m s−1
M1 for \(\frac{1440 + 900}{240 + 150}\) (distance in m and time in s)
A1 for 6 (m s−1)
Worked solution: Total distance \(1440 + 900 = 2340\) m; total time \(240 + 150 = 390\) s. \(\text{average speed} = \frac{2340}{390} = 6\text{ m s}^{-1}\)
(b)Answer: \(1.38\) m s−1 due east m s−1
M1 for the displacement \(1440 - 900 = 540\) m
M1 for \(\frac{\text{displacement}}{\text{total time}}\)
A1 for awrt 1.38 m s−1 due east
Worked solution: The displacement of \(B\) from \(A\) is \(1440 - 900 = 540\) m, i.e. 540 m due east. \(\text{average velocity} = \frac{540}{390} = 1.38\text{ m s}^{-1}\) due east.
(c)Answer: The cyclist changes direction, so the distance travelled (2340 m) is greater than the magnitude of the displacement (540 m), over the same time.
B1 for the distance travelled is greater than the magnitude of the displacement because the cyclist changes direction
Worked solution: Average speed uses the total distance, 2340 m, but average velocity uses the displacement, only 540 m, because part of the journey is back towards \(A\).
Question 3Hard10 marks
A particle \(P\) moves along a straight line with constant acceleration. At time \(t = 0\), \(P\) passes through the point \(O\), moving in the positive direction with speed \(u\) m s−1. At time \(t = 5\) seconds, \(P\) is at the point \(A\) and is moving in the negative direction with speed 8 m s−1. The total distance travelled by \(P\) in the interval \(0 \le t \le 5\) is 26 m.
(a) Explain why \(P\) is instantaneously at rest at some time during the interval \(0 \le t \le 5\).[1]
(b) Find the value of \(u\).[5]
(c) Find the acceleration of \(P\).[2]
(d) Find the displacement of \(A\) from \(O\).[2]
Show the answer and mark scheme
(a)Answer: The velocity changes from positive to negative. The acceleration is constant, so the velocity changes continuously (at a constant rate) and must be zero at some instant in between.
B1 for the velocity changes sign (positive to negative) and changes continuously, as the acceleration is constant, so it must pass through zero
Worked solution: At \(t = 0\) the velocity is positive and at \(t = 5\) it is negative. With constant acceleration the velocity changes at a constant rate, so it passes through zero: \(P\) is instantaneously at rest at that moment before moving back in the negative direction.
(b)Answer: \(u = 12\)
M1 for using the constant acceleration, e.g. \(|a| = \frac{u + 8}{5}\) or similar triangles on a velocity–time graph, to express the time at rest or the distances in terms of \(u\)
M1 for the total distance as the sum of the two distances, e.g. \(\frac{u^2}{2|a|} + \frac{8^2}{2|a|} = 26\)
A1 for a correct 3-term quadratic, e.g. \(5u^{2} - 52u - 96 = 0\)
dM1 for solving their quadratic
A1 for \(u = 12\) only, with the negative root rejected
Worked solution: The velocity–time graph is a straight line from \((0, u)\) to \((5, -8)\), so the magnitude of the acceleration is \(|a| = \frac{u + 8}{5}\). Distance while moving in the positive direction: \(\frac{u^2}{2|a|} = \frac{5u^2}{2(u + 8)}\); distance while moving in the negative direction: \(\frac{8^2}{2|a|} = \frac{320}{2(u + 8)}\). \(\frac{5(u^2 + 64)}{2(u + 8)} = 26 \Rightarrow 5(u^2 + 64) = 52(u + 8)\) \(\Rightarrow 5u^{2} - 52u - 96 = 0 \Rightarrow (u - 12)(5u + 8) = 0\), so \(u = 12\) or \(u = -\frac{8}{5}\). Since \(u\) is a speed, \(u = 12\).
(c)Answer: \(-4\) m s−2
M1 for \(\frac{v - u}{t} = \frac{-8 - 12}{5}\) using their \(u\)
A1 for \(-4\) (m s−2)
Worked solution: \(a = \frac{v - u}{t} = \frac{-8 - 12}{5} = -4\text{ m s}^{-2}\), i.e. 4 m s−2 in the negative direction.
(d)Answer: \(10\) m
M1 for \(s = \left(\frac{u + v}{2}\right)t = \frac{12 + (-8)}{2} \times 5\) or \(18 - 8\)
A1 for \(10\) (m)
Worked solution: \(s = \left(\frac{12 + (-8)}{2}\right) \times 5 = 10\) m, so \(A\) is 10 m from \(O\) in the positive direction (check: \(18 - 8 = 10\)).